COURSE NUMBER: ME 321

Fluid Mechanics I

3 credit hour

Basic Equations in fluid Dynamics

Course teacher

Dr. M. Mahbubur Razzaque

Professor

Department of Mechanical Engineering

BUET 1

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Description of Fluid Motion

In the description of a flow field, it is convenient to think of

individual particles each of which is considered to be a small mass of

fluid, consisting of a large number of molecules, that occupies a

small volume that moves with the flow.

If the fluid is incompressible, the volume does not change in

magnitude but may deform. If the fluid is compressible, as the

volume deforms, it also changes its magnitude. In both cases the

particles are considered to move through a flow field as an entity.

There are two approaches of mathematical description of the physical

quantities of a fluid motion as a function of space and time

coordinates :

Lagrangian Descriptions of Motion

and

Eulerian Descriptions of Motion

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Lagrangian Descriptions of Motion

In the study of particle mechanics, where attention is focused on

individual particles, motion is observed as a function of time. The

position, velocity, and acceleration of each particle are listed as s(x0,

y0, z0, t), V(x0, y0, z0, t), and a(x0, y0, z0, t), and quantities of interest

can be calculated.

The point (x0, y0, z0) locates the starting point—the name—of each

particle. This is the Lagrangian description, named after Joseph L.

Lagrange (1736–1813), of motion that is used in a course on

dynamics.

In the Lagrangian description many particles can be followed and

their influence on one another noted. This becomes, however, a

difficult task as the number of particles becomes extremely large in

even the simplest fluid flow.

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Eulerian Descriptions of Motion

The region of flow that is being considered is called a flow field.

In Eulerian description, points in a flow field is identified in space

and then the velocity of fluid particles passing each point is observed;

we can observe the rate of change of velocity as the particles pass

each point, that is,

and we can observe if the velocity is changing with time at each

particular point, that is, .

In this Eulerian description, named after Leonhard Euler (1707-

1783), of motion, the flow properties, such as velocity, are functions

of both space and time.

In Cartesian coordinates the velocity is expressed as V = V(x, y, z, t).

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Lagrangian and Eulerian Descriptions

An example may clarify these two ways of describing motion. An

engineering firm is hired to make recommendations that would

improve the traffic flow in a large city.

The engineering firm has two alternatives: Hire students to travel in

automobiles throughout the city recording the appropriate

observations (the Lagrangian approach), or

hire students to stand at the intersections and record the required

information (the Eulerian approach).

A correct interpretation of each set of data would lead to the same set

of recommendations, that is, the same solution. In an introductory

course in fluids, however, the Eulerian description is used

exclusively since the physical laws using the Eulerian description are

easier to apply to actual situations.

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Acceleration

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Acceleration

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Acceleration

This derivative is called the substantial derivative, or material

derivative. It is given a special name and special symbol (D/Dt

instead of d/dt) because we followed a particular fluid particle, that

is, we followed the substance (or material).

It represents the relationship between a Lagrangian derivative in

which a quantity depends on time t and an Eulerian derivative in

which a quantity depends on position (x, y, z) and time t.

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Acceleration

The time-derivative term on the right side of Eqs. 3.2.8 for the

acceleration is called the local acceleration and the remaining

terms on the right side in each equation form the convective

acceleration.

Hence the acceleration of a fluid particle is the sum of the local

acceleration and convective acceleration.

In a pipe, local acceleration results if, for example, a valve is being

opened or closed; and convective acceleration occurs in the

vicinity of a change in the pipe geometry, such as a pipe

contraction or an elbow. In both cases fluid particles change speed,

but for very different reasons.

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Basic Laws in Fluid Dynamics

The basic laws in fluid dynamics are expressed either in terms of a

system or a control volume.

A system is a fixed collection of material particles. For example, if

we consider flow through a pipe, we could identify a fixed quantity

of fluid at time t as the system (Fig. 4.1); this system would then

move due to velocity to a downstream location at time t + Dt. Any of

the basic laws could be applied to this system.

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Our interest is most often focused on a device, or a region of space,

into which fluid enters and/or from which fluid leaves; we identify

this region as a control volume.

An example of a fixed control volume is shown in Fig. 4.2a. A

control volume need not be fixed; it could deform. However, only

fixed control volumes will considered from now on.

The difference between a control volume and a system is illustrated

in Fig. 4.2b.

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System to control volume transformation

System to control-volume transformation, or equivalently, the

Reynolds transport theorem is written as,

Here h represents the intensive property (the property of the system

per unit mass) associated with Nsys. The relation between h and Nsys

is expressed as

The first integral of Eq. (4.2.9) represents the rate of change of the

extensive property in the control volume.

The second integral represents the flux of the extensive property

across the control surface; it may be nonzero only where fluid crosses

the control surface.

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CONSERVATION OF MASS

A system is a given collection of fluid particles; hence its mass

remains fixed:

Using Reynolds transport theorem and recognizing that Nsys = msys

i.e. the mass of the system (an extensive property for which h =1),

we can write

This is called the continuity equation. If the flow is steady, there

results

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CONSERVATION OF MASS

For a uniform flow with one entrance and one exit, Eq. (4.3.4) takes

the form

Where we have used

If the density is constant in the control volume, the continuity

equation reduces to

This form of the continuity equation is used quite often, particularly

with liquids and low-speed gas flows.

The above derivation assumed uniform velocities at the inlet and exit

sections. What happens when the velocities at the inlet and the exit

sections are non uniform?

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Continuity equation

Suppose that the velocity profiles at the entrance and the exit are not

uniform, such as sketched in Fig. 4.7. Furthermore, suppose that the

density is uniform over each area. Then the continuity equation takes

the form

Where, are the average velocities at sections I and 2,

respectively. In examples and problems the overbar is often omitted.

It should be kept in mind, however, that actual velocity profiles are

usually not uniform.

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Continuity equation

Two fluxes are useful in specifying the quantity of flow. The mass

flux or the mass rate of flow is

and has units of kg/s; Vn is the normal component of velocity. The

flow rate Q, the volume rate of flow, is

and has units of m3/s. The mass flux is usually used in specifying the

quantity of flow for a compressible flow and the flow rate for an

incompressible flow.

In terms of average velocity, we have

Where uniform density and normal velocity is assumed.

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Solution

The control volume is selected to be the inside of the nozzle as shown. Flow

enters the control volume at section 1 and leaves at section 2. The

simplified continuity equation (4.4.6) is used since the density of water is

assumed constant and the velocity profiles are uniform:

The flow rate or discharge is found to be:

Example 4.1

Water flows at a uniform velocity of 3 m/s into a nozzle that reduces the

diameter from 10 cm to 2 cm. Calculate the water’s velocity leaving the

nozzle and the flow rate.

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or

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Continuity equation in differential form

Consider the mass flux through each face of the infinitesimal control

volume shown in Fig. 5.1. We set the net flux of mass entering the

element equal to the rate of change of the mass of the element; that is,

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This takes the form

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Then the continuity equation becomes,

Or, in vector form,

The continuity equation in cylindrical coordinate is

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These are the called Navier-Stokes equations, named after C. I. M.

H. Navier (l785-1836) and Sir George G. Stokes (1819-1903), who

are credited with their derivation. These are second-order nonlinear

partial differential equations.

Navier-Stokes equations

In an incompressible flow the density remains constant or the change

in density is very negligible. In an incompressible flow of a

Newtonian Fluid with constant viscosity, the momentum equations

(derived from Newton’s second law of motion) are:

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Which may be interpreted as:

Inertia force per unit vol. = gravity force per unit vol. + pressure

force per unit vol. + viscous force per unit volume

There are four unknowns: p, u, v, and w. So, for solution, it should be

combined with the incompressible continuity relation to form four

equations in these four unknowns.

The continuity equation for incompressible flow in steady state is:

Or,

Navier-Stokes equations ….

In vector notations, the Navier-Stokes equations may be written as:

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Incompressible Inviscid flow….

Computation Fluid Dynamics (CFD) is the branch of fluid

mechanics that uses numerical methods to solve the Navier-Stokes

equations and continuity equation with appropriate boundary

conditions and without any analytical simplification.

However, analytical solution for many viscous flow problems are

also possible through some simplifying assumptions.

If viscosity is assumed to be zero (inviscid flow approximation), the

second order terms in the Navier-Stokes equations are lost and we

get the Euler’s Equation, as:

Integration of this equation along a streamline results in Bernoulli’s equation.

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Energy Equation

Many problems involving fluid motion demand that the energy

equation, be used to relate quantities of interest.

If the heat transferred to a device (a boiler or compressor), or the work

done by a device (a pump or turbine), is desired, the energy equation

is needed.

It is also used to relate pressures and velocities when Bernoulli’s

equation is not applicable; this is the case whenever viscous effects

cannot be neglected, such as flow through a piping system.

Let us express the energy equation in control volume form. For a

system it is

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where the specific energy e is,

In terms of a control volume, Eq. 4.5.1 becomes

This can be put in simplified forms for certain restricted flows. The

rate-of-heat transfer term ˙Q represents the rate-of-energy transfer

across the control surface due to a temperature difference. (Do not

confuse this term with the flow rate Q.)

The work-rate term results from work being done by the system. Work

of interest in fluid mechanics is due to a force moving through a

distance while it acts on the control volume. The The work-rate term

is discussed in detail in the following section.

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Considering different situations in fluid problems, the work-rate term

becomes:

We should note that third and fourth components in the work-rate

term are seldom encountered in problems in an introductory course

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In many fluid flows, useful forms of energy (kinetic energy and potential

energy) and flow work are converted into unusable energy forms (internal

energy or heat transfer). However, in an introductory fluid mechanics

course the sum of these effects is lumped together. We define losses as the

sum of all the terms representing unusable forms of energy:

For a steady-flow situation in which there is one entrance and one exit

across which uniform profiles can be assumed, the energy equation

becomes.

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Here, Ws is shaft work and hL is head loss. If the losses are negligible,

if there is no shaft work or heat flow and if the flow is incompressible,

we can write the energy equation in its most useful form as,

Note that this form of energy equation is identical with Bernoulli's

equation. However, we must understand that the Bernoulli's equation

is a momentum equation applicable along a stream line and the

equation above is an energy equation applied between two sections of

a flow.

When the energy equation is applied to any steady uniform flow, the

control volume is usually selected such that the entrance and exit

sections have a uniform total head.

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For example, if the energy equation is applied to a flow passing a

gate, the appropriate control volume may be chosen as shown below

The total head at the entrance and exit can be evaluated at an point at

the entrance and exit, respectively. However, a convenient choice

would be the points at the water surface. Thus the energy equation

becomes

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where we have introduced the head loss hL, often written in terms of a

loss coefficient K as

A final note to this section regards nomenclature for pumps and

turbines in a flow system. It is often conventional to call the energy

term associated with a pump the pump head Hp, and the term

associated with a turbine the turbine head HT . Then the energy

equation takes the form

In this form we have equated the energy at the inlet plus added energy

to the energy at the exit plus extracted energy (energy per unit weight,

of course). If any of the quantities is zero (e.g., there is no pump), the

appropriate term is simply omitted.

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The power generated by a turbine with an efficiency of hT is

The power required by a pump with an efficiency of hp would be

Usually power in expressed in kilowatts or horsepower. One

horsepower is equivalent to 0.746 kW.

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Momentum Equation

Newton's second law, often called the momentum equation, states

that the resultant force acting on a system equals the rate of change of

momentum of the system. If a device has entrances and exits across

which the velocity is assumed to be uniform and if the flow is steady,

the momentum equation can then be written as

Where,

If we consider the nozzle

shown and determine the x-

component of the force of

the joint on the nozzle,

the momentum equation x-

direction becomes

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An example of a free-surface flow in a rectangular channel is shown

in Fig. 4.12. If we want to determine the force of the gate on the flow,

the following expression can be derived from the momentum

equation:

where F1 and F2 are pressure forces.

Momentum Equation Applied to Deflectors

The application of the momentum equation to deflectors forms an

integral part of the analysis of many turbomachines, such as turbines

and pumps.

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Momentum Equation Applied to Deflectors

The analysis is separated into two parts: jets deflected by stationary

deflectors and jets deflected by moving deflectors. For both problems

we will assume the following:

-The pressure external to the jets is everywhere constant so that the

pressure in the fluid entering the deflector is the same as that in the

fluid exiting the deflector.

-The frictional resistance due to the fluid-deflector interaction is

negligible so that the relative speed between the deflector surface and

the jet stream remains unchanged.

- Lateral spreading of a plane jet is neglected.

-Body forces are small and will be neglected.

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Stationary Deflector

Let us first consider the stationary deflector, illustrated in Fig. 4.13.

Bernoulli's equation allows us to conclude that V2 = V1 since the

pressure is assumed to be constant external to the fluid jet and

elevation changes are negligible.

Assuming steady, uniform flow the x- and y- components of the

momentum equation takes the form of Eq. 4.5.15.

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For given jet conditions the reactive force components can be

calculated.

Moving Deflectors

The situation involving a moving deflector depends on whether a

single deflector is moving (a water scoop used to slow a high-speed

train) or whether a series of deflectors is moving (the vanes on a

turbine).

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Moving Deflectors

Let us first consider that the single deflector shown in Fig. 4.14 is

moving in the positive x-direction with the speed VB. In a reference

frame attached to the stationary nozzle, from which the fluid jet

issues, the flow is unsteady; that is, at a particular point in space, the

flow situation varies with time.

A steady flow is observed, however, from a reference frame attached

the deflector.

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From this reference frame, moving with the constant velocity VB, we

observe the relative speed Vr entering the control volume to be V1 -

VB, as shown in Fig. 4.14. It is this speed that remains constant as the

fluid flows relative to the deflector. Hence the momentum equation

takes the forms

Where represents only that part of the mass flux exiting the fixed

jet that has its momentum changed.

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Since the deflector moves away from the jet some of the fluid that

exits the fixed jet never experiences a momentum change; this fluid is

represented by the distance VB Dt, shown in Fig. 4.14. Hence

Where the relative speed (V1 - VB) is used in the calculation; the mass

flux rAVB is subtracted from the exiting mass flux pAV1 to provide

the mass flux .

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Series of vanes

For a series of vanes (a cascade) the jets may be oriented to the side,

as shown in Fig. 4.15. The actual force on a particular vane would be

zero until the jet strikes the vane; then the force would increase to a

maximum and decrease to zero as the vane leaves the jet.

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Assume that, on the average, the jet is deflected by the vanes as

shown as viewed from a stationary reference frame; the fluid jet enters

the vanes with an angle b1 and exits with an angle b2. What is desired,

however, is that the relative velocity enter the vanes tangent to the

leading edge of the vanes, that is, Vr1 in Fig. 4.16b is at the angle a1.

The relative speed then remains constant as it travels over the vane

with the exiting relative velocity Vr2 leaving with the vane angle

a2.The relative and absolute velocities are related with the velocity

polygons of Fig. 4.16 b and c.

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Assuming that all of the mass exiting the fixed jet has its momentum

changed, we can write the momentum equation as

Interest is usually focused on the x-component of force since it is this

component that is related to the power output (or requirement). The

power would be found by multiplying the x-component force by the

blade speed for each jet; this takes the form

where N represents the number of jets.

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EXAMPLE 4.11

Water flows through a horizontal pipe bend and exits into the

atmosphere. The flow rate is 0.01 m3/s. Calculate the force in each of

the rods holding the pipe bend in position. Neglect body forces and

viscous effects.

Solution

We have selected a control volume that surrounds the bend as shown.

Since the rods have been cut, the forces that the rods exert on the

control volume are included. The pressure forces at the entrance and

exit of the control volume are also shown.

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The flexible section is capable of resisting the interior pressure but it

transmits no axial force or moment. The body force (weight of the

control volume does not act in the x- or y- direction but normal to it.

Therefore, no other forces are shown. The average velocities are

found to be

Before we can calculate the forces Rx and Ry we need to find the

pressures p1 and p2. The pressure p2 is zero because the flow exits into

the atmosphere. The pressure at section 1 can be determined using the

energy equation or the Bernoulli equation.

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Neglecting losses between sections 1 and 2, the energy equation gives

Now we can apply the momentum equation in the x-direction to find

Rx and in the y- direction to find Ry:

Note that we have assumed uniform profiles and steady flow. These

are the usual assumptions if information is not given otherwise.

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EXAMPLE 4.12

A deflector turns a sheet of water through an angle of 30o as shown.

What force per unit width is necessary to hold the deflector in place if

the mass flux is 32kg/s?

Solution

The control volume we have selected includes the deflector and the

water adjacent to it. The only force that is acting on the control

volume is due to the support. This force has been decomposed into Rx

and Ry.

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The velocity V1 is found to be

Bernoulli’s equation shows that if the pressure does not change, then

the magnitude of the velocity does not change, provided that there is

no significant change in elevation and that viscous effects are

negligible; thus we can conclude that V2 = V1 since p2 = p1.

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Next, the momentum equation is applied in the x-direction to find Rx

and then in the y-direction for Ry:

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EXAMPLE 4.13

The deflector shown moves to the right at 3 m/s while the nozzle

remains stationary. Determine (a) the force components needed to

move the deflector, (b) V2 as observed from a fixed observer, and (c)

the power generated by the vane. The jet velocity is 8 m/s.

Solution

(a) To solve the problem of a moving deflector, we will observe the

flow from a reference frame attached to the deflector. In this moving

reference frame the flow is steady and Bernoulli's equation can then

be used to show that Vr1 = Vr2 = 5 m/s, the velocity of the sheet of

water as observed from the deflector. Note that we cannot apply

Bernoulli's equation in a fixed reference frame since the flow would

not be steady.

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Applying the momentum equation to the moving control volume,

which is indicated again by the dashed line, we obtain the following:

When calculating the mass flux, we must use only that water which

has its momentum changed; hence the velocity used is 5 m/s.

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(b) Observed from a fixed observer the velocity V2 of the fluid after

the deflection is V2 = Vr2 + VB, where Vr2 is directed tangential to the

deflector at the exit and has a magnitude equal to Vr1 (see the velocity

diagram above). Thus

Finally,

(c) The power generated by the moving vane is equal to the velocity

of the vane times the force the vane exerts in the direction of the

motion. Therefore,

= 3 x 26.8 = 80.4 W

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EXAMPLE 4.17

Find an expression for the head loss in a sudden expansion in a

pipe in terms of V1 and the area ratio. Assume uniform velocity

profiles and assume that the pressure at the sudden enlargement

is p1.

Solution

A sketch is shown of a sudden expansion with the diameter

changing from d1 to d2. The pressure at the sudden enlargement

is p1, so that the force acting on the left end shown is p1A2.

Newton's second law applied to the control volume yields,

assuming uniform profiles,

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The energy equation provides

To express this in terms of only V1, we can use continuity and

relate

Then the expression above for the head loss becomes